Question

The altitude of a triangle is increasing at a rate of 2.500 centimeters/minute while the area of the triangle is increasing at a rate of 1.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.500 centimeters and the area is 93.000 square centimeters?

Answer 1

Given
A=bh/2, b=2*93/10.5=17.714
differentiate with respect to t (by the product rule on the right hand side).
The only unknown left is the rate of change of the base.
Hint: it is negative.

Answer 2

let the area be A
let the base be x
let the height be y

given: at a time of t minutes,
dA/dt = 1.5 cm^2/min
dy/dt = 2.5 cm^2/min

find:
dx/dt when A = 93 and y = 10.5

A = xy/2 or
2A = xy (equ#1)

differentiate implicitly with respect to t, using the product rule
2dA/dt =x(dy/dt) + y(dx/dt) (equ#2)

we know A=93 when y = 10.5, so x = 17.714
sub into equ#2
2(1.5) = 17.714(2.5) + 10.5dx/dt

solve for dx/dt

Answer 3

My answer is -3.9319, but it's wrong.

Answer 4

I have -3.932 too.
Can you check the numbers in the question?

Answer 5

I copied & pasted this from my online homework. It said the answer is wrong.

Answer 6

I also had -3.932

What answer did the book have?

Answer 7

I use an online homework program called WebWork. It just tells you if your answer is correct or not.

Answer 8

Try -3.93, -3.932.

Answer 9

The answer is -3.9300. Thanks!